This is a follow-up question to Rate of convergence of $\frac{1}{\sqrt{n\ln n}}(\sum_{k=1}^n 1/\sqrt{X_k}-2n)$, $X_i$ i.i.d. uniform on $[0,1]$? . My motivation is to construct a statistic whose rate convergence to a Gaussian will be very slow and as such explore types of convergence which are not encapsulated by the Berry-Esseen' theorem. We therefore define the following statistic: \begin{equation} S_n := \frac{\left(\sum\limits_{k=1}^n f^{-1}(X_k) - \frac{3}{2} e \cdot n\right)}{e\cdot \sqrt{n \log(\log(n))}} \end{equation} where $X_k$ are i.i.d. uniformly distributed in $(0,1)$ and the function $f()$ is defined as follows: \begin{equation} f(x) := \frac{e^2}{2} \cdot \frac{1+\log(x)}{x^2 \log(x)^2} 1_{x \ge e} \end{equation} Now, the probability density of $f(X)$ is as follows: \begin{eqnarray} \rho_{f^{-1}(X)}(z) &=& \int\limits_0^1 \delta(z - f^{-1}(x)) dx =-\int\limits_{e}^\infty \delta(z-u) f^{'}(u) du= -f^{'}(z) 1_{z \ge e}\\ &=& \frac{e^2}{2} \cdot \frac{2+3 \log(z)+2 \log(z)^2}{z^3 \log(z)^3}1_{z \ge e} \end{eqnarray} From this we readily get the moments: \begin{eqnarray} E\left[ f^{-1}(X) \right] = \frac{3}{2} e\\ E\left[ (f^{-1}(X))^2 \right] = \infty \end{eqnarray} We also get the characteristic function. It reads: \begin{eqnarray} \kappa_{f^{-1}(X)}(k) = e^{\imath k e}+ \imath k \frac{e}{2} e^{\imath k e}-k^2 \frac{e^2}{2}\cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta \end{eqnarray} for $0<k<1$.

Note: The last integral on the right hand side is for me hard to crack. However numerical computations suggest that: \begin{equation} \lim_{k\rightarrow 0} \frac{1}{\log(\log(1/k))} \cdot \int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta = 1 \end{equation} Indeed by using the integral representation of the Gamma function along with integration by parts we quickly establish the following identity: \begin{eqnarray} (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) = \frac{e^{-\delta}}{\delta} + (-\imath k)^\delta \cdot \Gamma(-\delta) + \sum\limits_{n=1}^\infty \frac{(\imath k)^n}{n!}\cdot \frac{e^{n-\delta}}{\delta-n} \end{eqnarray}

Now clearly \begin{eqnarray} &&\int\limits_0^\infty (-\imath k)^\delta \cdot \Gamma(-\delta,-\imath e k) d \delta =\\ && \int\limits_0^\infty \left( \frac{e^{-\delta}}{\delta} + (-\imath k)^\delta \cdot \Gamma(-\delta) \right) d\delta + O(k)\\ &&= \int\limits_0^\infty \left( \frac{e^{-\delta}}{\delta} - \frac{(-\imath k)^\delta}{\delta} \right) d\delta + \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k)\\ &&= \left.\left( Ei(-\delta) - Ei(-A \delta)\right)\right|_0^\infty+ \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k)\\ &&= \log(-A) + \int\limits_0^\infty (-\imath k)^\delta \left(\Gamma(-\delta)+\frac{1}{\delta}\right) d\delta + O(k) \end{eqnarray} where $A=-\log(-\imath k)= \imath \pi/2 - \log(k)$. Now we have checked numericaly that the integral in the middle above decays monotonically when $k\rightarrow 0$. Since now $\log(-A) = \log(-\imath \pi/2 + \log(k))= \log(-\imath \pi/2-\log(1/k)) \rightarrow \log(-\log(1/k)) = -\imath \pi/2 + \log(\log(1/k)) \rightarrow \log(\log(1/k))$ when $k\rightarrow 0$ the claim is established.

Now we check that our test statistic is properly normalized.

Define $c_n:=\sqrt{n\log(\log(n))}$. Indeed we have: \begin{eqnarray} &&\log\left( \kappa_{S_n}(k)\right) =\\ && -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ \kappa_{f^{-1}(X)}(\frac{k}{e c_n})\right] \\ &&= -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ e^{\imath \frac{k}{c_n}}(1+\imath \frac{k}{2 c_n}) - \frac{1}{2} \frac{k^2}{c_n^2} \log(1+\log(c_n)-\log(k))\right]\\ &&= -\imath k \frac{3}{2} \frac{n}{c_n} + n \log\left[ 1+\imath \frac{3}{2} \frac{k}{c_n} - \frac{k^2}{c_n^2} + O(\frac{k^3}{c_n^3}) - \frac{1}{2} \frac{k^2}{c_n^2} \log(1+\log(c_n)-\log(k)) \right]\\ &&= -\imath k \frac{3}{2} \frac{n}{c_n} + \imath k \frac{3}{2} \frac{n}{c_n} + \left(\frac{1}{8} - \frac{1}{2} \log(1+\log(c_n)-\log(k))\right) \frac{k^2}{c_n^2} n + O(\frac{k^3}{c_n^3})\\ &&= \left(\frac{1}{8} - \frac{1}{2} \log(1+\log(c_n)-\log(k))\right) \frac{k^2}{c_n^2} n + O(\frac{k^3}{c_n^3}) \end{eqnarray} Now for the statistic to be properly normalized we have to have: \begin{equation} \lim_{n\rightarrow \infty} \frac{n}{c_n^2} \log(\log(c_n)) = 1 \end{equation} which is indeed the case as one can readily check by plugging the definition of $c_n$ into the lhs.

Now, I carried out a Monte Carlo simulation and computed the sample Cumulative Distribution Function (CDF) of our statistic and plotted it along with the CDF of a standardized Gaussian distribution with the former and the later being plotted in Blue and Purple respectively. Here I took $n=5,10,15$ and in each case I used $m=1000$ realizations. The figures are below:

I have used the following Mathematica code to produce those figures:

```
m = 1000; n = 15; delta = 1/10;
bins = Table[-5 + delta/2 + j delta, {j, 1, (10 - delta)/delta}];
limD = CDF[NormalDistribution[0, 1], bins];
X = RandomReal[{0, 1}, {m, n}];
x =.; {t0, Y} =
Timing[(x /.
Map[First[
NSolve[(E^2 (1 + Log[x]))/(2 x^2 Log[x]^2) == # && x > E, x,
Reals]] &, X, {2}])];
ll = (Total[#] & /@ Y - 3/2 E n)/(E Sqrt[n Log[Log[n]]];
emp = EmpiricalDistribution[ll];
DD = CDF[emp, bins];
pl = ListPlot[Transpose[{bins, #}] & /@ {DD, limD}, ImageSize -> 800,
LabelStyle -> {15, FontFamily -> "Arial"},
BaseStyle -> {15, FontFamily -> "Bold"},
PlotLabel -> "n=" <> ToString[n]];
Export["LimitBehavior1_n_" <> ToString[n] <> ".jpg", pl, "JPEG"];
Import["LimitBehavior1_n_" <> ToString[n] <> ".jpg"]
```

Having said all this my question is the following. What is the rate of convergence of our statistic towards a Gaussian. To be specific we are asking about the behavior of the supremum norm of the difference in CDFs for large values of $n$.

With this normalization factor, the source distributions the statistic works over vary with the statistic's index $n$.The Berry-Esseen theorem doesn't make a statement about this situation. $\endgroup$ – enthdegree Apr 5 '18 at 18:38